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About the transcendentality of some numbers
Jamel Ghannouchi
To cite this version:
Jamel Ghannouchi. About the transcendentality of some numbers. 2013. �hal-00808463v11�
About the transcendentality of some numbers
Jamel Ghanouchi
Ecole Supérieure des Sciences et Techniques de Tunis jamel.ghanouchi@topnet.tn
Abstract
Everyone knows that π, e or Cn, the Champernowne number are trans- cendentals, but what about π + e, π + Cn or e+ Cn? In this paper, we demonstrate a method in order to know if they also are.
The approach
A number is transcendantal if it is not the root of a polynomial equation anxn+an−1xn−1+...+a0 = 0
where ai are rational and different from zero, otherwise it would be al- gebraic. We know thatπ, eand Cn, the Champernowne number are tran- sendentals, but we do not know anything aboute+π, π +Cn ore+Cn. Effectively, ifAis transcendental :B transcendantal, we still do not know the nature ofA+BorA−B. But ifBis algebraic, thenA+B andA−B are transcendantals. And ifAandB are algebraics, their sum as well as their différence are algebraics. Let us trie to solve this problem. Let us take
C1 =
1 2 3 4 5 4 5 1 2 3 2 3 4 5 1 5 1 2 3 4 3 4 5 1 2
And take
C′1 =
0 5 10 15 20 10 15 20 0 5 20 0 5 10 15
5 10 15 20 0 15 20 0 5 10
ThusC1+C′1 is a magic square. It contains all the numbers from1to25.
Let us take
A=
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
And
C2 =
C1 C1+ 25A C1+ 50A C1+ 75A C1+ 100A C1+ 75A C1+ 100A C1 C1+ 25A C1+ 50A C1+ 25A C1+ 50A C1+ 75A C1+ 100A C1 C1+ 100A C1 C1+ 25A C1+ 50A C1+ 75A
C1+ 50A C1+ 75A C1+ 100A C1 C1+ 25A
And
C′2 =
C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1 C′1
The mathematical induction is
Ci+1 =
Ci Ci+ 5iA Ci+ 2.5iA Ci+ 3.5iA Ci+ 4.5iA C1+ 3.5iA C1+ 4.5iA C1 C1 + 5iA C1+ 2.5iA
C1+ 5iA C1+ 2.5iA C1+ 3.5iA C1+ 4.5iA C1 C1+ 4.5iA C1 C1+ 5iA C1+ 2.5iA C1+ 3.5iA C1+ 2.5iA C1+ 3.5iA C1+ 4.5iA C1 C1+ 5iA
And
C′i+1 =
C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i C′i
AndCi+C′i is a magic square containing all the numbers from1to 5 + 20(5 + 52+...5i−1) + 20 = 5 + 100(1 + 5 + 52+...+ 5i−2) + 20
= 5 + 100(5i−1−1
4 ) + 20 = 5 + 25(5i−1−1) + 20 = 5i+1
The sum ofCi+1 isSi+1 = 5Si+ 10.5i+ 5S′i(S1 = 65,S′i = 50.5i−1). Hence Si+1 = 5Si+ 60.5i
5Si = 52Si−1 + 60.5i
...
5i−1S2 = 5iS1+ 60.5i 5iS1 = 5i(65) = 13.5i+1 Si+1 = (60i+ 65)5i = (12i+ 13)5i+1
If M is the greatest number of the square, S the sum, the value of the integerπ.10pi that approachesπin the square (example between1and25, π.10pi = 3and between1and125, it is31). The numberπ(pi) depends on M and S. This square contains pi first digit ofπ without the dot, we note it : π.10pi. It contains also e.10pi and Cn.10pi (and also π +e, πe, π+Cn, πCn...). Fori+ 1, the greatest number in the square isM = 5i+2which has E((i+2) log 5
log 10 ) =i+ 1ciphers and begins with 1,2,3and thenpi =i, or with a greater number and then pi = i+ 1. The sum is Si+1 = (12i + 13)5i+1 which has less thanpi+ 2 ciphers. The sumSi.10−pi tends, in the infinity, to a numberSwhich has less than three ciphers ! ButSi is rational, thusS is rational and algebraic.
Main results
S = 20 A=S−π B =S−e C =S−Cn A1 =S−π−e A2 =S−π+e A3 =S−πe A4 =S−π2 We know that
S =π+A=e+B =π+e+A1 =π−e+A2 =πe+A3
Definition The following numbers can be called bricks or elements, be- cause they constitute the bricks of the numbers. They exist, of course, and we prefer to call them prime numbers because they really generalize the concept of primes. Let us see this : A real number is compound if it is equal to±pn11...pini wherepj are integer prime numbers andnj are rationals. We define other real prime numbers which cannot be expressed like this :π,e, ln (2). Thus√q p=p
1
q is compound.
Also √np+ 1is prime, with pprime, hence√p−1 = (p−1)(√p+ 1)−1 is compound !
Furthermoreπandeare primes instead ofπn0 andem0 with(n0−1)(m0− 1)6= 0.
We define the GCD of two numbers as following : If p1 and p2 are prime real numbers
p1 6=p2 ⇒GCD(p1, p2) = 1 n1n2 <0⇒GCD(pn11, pn12) = 1 n1n2 >0⇒GCd(p1n1, pn12) =pmin(n1 1,n2) GCD(pn11pn22...pnii, p1′n′1p2′n′2...pj′n′j) =Y
i,j
GCD(pnii, p′jn′j) And ifx=pn11pn22...pini andy=pml1l1...pml lj
i thenydividesxif1< lk< iand forli =j,njmli >0,|mli|<|nj|.
Thus 32 does not divise the prime3.
Theorem IfT1′ and T2′ are transcendental prime real numbers, T1′T2′ and T1′+T2′ are transcendentals.
Proof of the theorem Let us takeT1andT2 two real compound transcen- dental numbers.
We have4possibilities 1)
T1T2nandT1+mT2are algebraics 2)
T1T2nis algebraic andT1+mT2 is transcendental 3)
T1T2nis transcendental andT1+mT2 is algebraic 4)
T1T2nandT1+mT2are transcendentals,∀m, n Thus
1)
T1T2n =AAandT1+mT2 =A′Aare algebraics, then T1T2n+mT2n+1 =AA+mT2n+1 =A′AT2n
AndT2 is supposed to be the fitting solution to this algebraic equation : it is impossible !
2)
IfT1T2nandT1T2n′ are algebraics, then
T1T2n(T1T2n′)−1 =T2n−n′ is algebraic and it is impossible withn 6=n′
There is only one n = n0 for which T1T2n is algebraic, all the others are transcendentals !
If we suppose thatn0is not unique, there are three possibilities :T1T1′(T2T2′)n is transcendental for alln orT1T1′−1(T2T2′−1)nis transcendental for alln or
there existsl0 andl′0 for which
T1T2n0 =AA T1′T2′n′0 =A′A T1T1′(T2′T2)l0 =A′′A T1T1′−1(T2T2′−1)l′0 =A′′′A are algebraics, we have
AA2T2l0+l0′−2n0T2′l0−l′0 =A′′AA′′′A AA′2T2l0−l′0T2′l0+l0′−2n′0 =A′′AA′′′−1A AAA′AT2l0−n0T2′l0−n′0 =A′′A AAA′−1A T2l0′−n0T2n′0−l0′ =A′′A
⇒
T2A=A′T2′B A=l0+l′0−2n0 B =l0′ −l0 A′ =A′′AA′′′AA−2A T2−B =B′T2′C
C =−(l0+l0−2n′0) B′ =A′2AA′′−A 1A′′′A T2D =C′T2′E D=l0−n0 E =n′0−l0 C′ =A′′AA−1A A′−1A T2F =D′T2′G F =l′0−n0 G=l′0−n′0 D′ =A′′′AA−1A A′A
⇒
T2−AB =A′−BT2′−B2
=T2−AB =B′AT2′AC T2′BC =A′−CT2AC
=T2′BC =B′−BT2−B2 T2DG=C′GT2′EG
T2EF =D′ET2′EG =D′EC′−GT2DG T2AE =A′ET2′BE =T2DBC′−B A=l0+l′0−2n0
B =l0′ −l0 A′ =A′′AA′′′AA−2A C =−(l0 +l0−2n′0) B′ =A′2AA′′−1A A′′′A D=l0−n0 E =n′0−l0 F =l′0−n0 G=l′0−n′0 C′ =A′′AA−A1A′−A1 D′ =A′′′AA−1A A′A
⇒
−B2 =AC=−(l0−l0′)2 =−(l0+l′0−2n0)(l0+l0′ −2n′0) EF =DG= (n′0−l0)(l′0−n0) = (l0−n0)(l′0−n′0)
AE =DB = (l0+l0′ −2n′0)(n0′ −l0) = (l′0−l0)(l0−n0) B′−B2 =B′AC =A′−BC
A =l0+l0′ −2n0 B =l′0−l0 A′ =A′′AA′′′AA−2A C =−(l0+l0−2n′0) B′ =A′2AA′′−A 1A′′′A D =l0 −n0 E =n′0−l0 F =l0′ −n0 G=l0′ −n′0 C′ =A′′AA−1A A′−1A D′ =A′′′AA−A1A′A
⇒
−B2 =AC =−(l0−l0′)2 =−(l0+l′0−2n0)(l0+l′0−2n′0) (n0−n′0)2 = (n0+n′0−2l0)(n0 +n′0−2l′0)
(n′0−l0)(l′0−n0) = (l0−n0)(l0′ −n′0)
AE =DB = (l0+l′0−2n′0)(n0′ −l0) = (l0′ −l0)(l0−n0) (n′0−l0)(l0+ 2l′0−n0−2n′0) = (l0−n0)(2l0′ −l0−n′0)
= (n′0−l0)(l0−n0) + (n′0−l0)(2l′0−2n′0) = (l0−n0)(l0′ −n′0) + (l0−n0)(l′0−l0) A=l0+l0′ −2n0
B =l0′ −l0 A′ =A′′AA′′′AA−2A C =−(l0+l0−2n′0) B′ =A′2AA′′−1A A′′′A D=l0−n0 E =n′0−l0 F =l′0−n0 G=l′0−n′0 C′ =A′′AA−A1A′−A1 D′ =A′′′AA−1A A′A
⇒
−B2 =AC =−(l0−l0′)2 =−(l0+l′0−2n0)(l0+l′0−2n′0) (n0−n′0)2 = (n0+n′0−2l0)(n0 +n′0−2l′0)
(n′0−l0)(l′0−n0) = (l0−n0)(l0′ −n′0)
AE =DB = (l0+l′0−2n′0)(n0′ −l0) = (l0′ −l0)(l0−n0)
(l0−n0)(n′0−l0′) = (l′0−n′0)(l0 −n0−2n′0+ 2l0) = (l′0−n′0)(3l0−n0−2n′0) A=l0+l0′ −2n0
B =l0′ −l0 A′ =A′′AA′′′AA−2A C =−(l0+l0−2n′0) B′ =A′2AA′′−1A A′′′A D=l0−n0 E =n′0−l0 F =l′0−n0 G=l′0−n′0 C′ =A′′AA−1A A′−1A D′ =A′′′AA−1A A′A
⇒
−B2 =AC =−(l0−l0′)2 =−(l0+l′0−2n0)(l0+l′0−2n′0) (n0−n′0)2 = (n0+n′0−2l0)(n0 +n′0−2l′0)
(n′0−l0)(l′0−n0) = (l0−n0)(l0′ −n′0)
AE =DB = (l0+l′0−2n′0)(n0′ −l0) = (l0′ −l0)(l0−n0)
(l0′ −n′0)(−n0+l0+ 3l0−n0−2n′0) = 0 = (l0′ −n′0)(4l0−2n0−2n′0)A=l0+l0′ −2n0 B =l0′ −l0
A′ =A′′AA′′′AA−A2 C =−(l0+l0−2n′0) B′ =A′A2A′′−1A A′′′A D=l0−n0 E =n′0−l0 F =l′0−n0 G=l′0−n′0 C′ =A′′AA−1A A′−1A D′ =A′′′AA−1A A′A
⇒l0 =l′0 =n′0 =n0 n0 is unique
3)
(T1 +nT2)−(T1 +n′T2) = (n−n′)T2 = AAis algebraic,T2 could not be a solution to this algebraic equation, this can not be possible whilen=n′. There is only one possibility, it is whenn=m0makingT1+nT2 algebraic, since all the others are transcendentals !
If m0 is not unique, there are three possibilities : T1 +T1′ +m(T2 +T2′) is transcendental for allmorT1−T1′+m(T2−T2′)is transcendental for allm or there existsl0, l′0 for which
T1+m0T2 =AA T1′+m′0T2′ =A′A
T1+T1′ +l0(T2+T2′) =A′′A T1−T1′ +l′0(T2−T2′) = A′′′A algebraics, then
2T1+l0(T2+T2′) +l0′(T2−T2′) = A′′A+A′′′A
= 2T1+ 2m0T2+l0(T2+T2′) +l0′(T2−T2′)−2m0T2
= 2AA+ (l0+l′0−2m0)T2 + (l0−l0′)T2′ 2T1′ +l0(T2+T2′)−l′0(T2−T2′) =A′′A−A′′′A 2A′A+ (l0 −l′0)T2+ (l0+l′0−2m′0)T2′
⇒
AT2+BT2′ =C A=l0+l′0−2m0
B =l0−l′0
C =A′′A+A′′′A−2AA
BT2+B′T2′ =C′ B′ =l0+l′0−2m′0 C′ =A′′A−A′′′A −2A′A
⇒
ABT2+B2T2′ =BC B′AT2+BB′T2′ =B′C ABT2+AB′T2′ =AC′ B2T2+BB′T2′ =B′C′ A=l0+l′0−2m0 B =l′0−l0
C=A′′A+A′′′A−2AA
B′ =l0+l′0−2m′0 C′ =A′′A−A′′′A−2A′A
⇒
B2 =AB′
ABT2+AB′T2′ =BC =AC′ BC =AC′
B′C′ =B′C A=l0+l′0−2m0
B =l′0−l0
C =A′′A+A′′′A−2AA
B′ =l0+l0′ −2m′0 C′ =A′′A−A′′′A−2A′A
⇒
(l0−l′0)2 = (l0 +l0′ −2m0)(l0+l0′ −2m′0) (l0′ −l0)C= (l0+l′0−2m0)C′
(l0+l0′ −2m′0)(C′−C) = 0 A =l0+l0′ −2m0
B =l0′ −l0
C =A′′A+A′′′A−2AA B′ =l0+l′0−2m′0 C′ =A′′A−A′′′A −2A′A
⇒l0 =l′0 =m0 =m′0 m0 is unique !
Thus there are finally two possibilities : I)
There are three sub-possibilities :
T1T1′′(T2T2′′)n is transcendental for alln, for allT1′, T2′ transcendental prime numbers, there existsT1, T1′′, T2, T2′′for whichT1 =T1′T1′′−1andT2 =T2′T2′′−1, in this caseT1′T2′ is always transcendental and of courseT1′+T2′ too.
IfT1T1′′−1(T2T2′′−1)nis transcendental for alln, for allT1′, T2′ transcendental prime numbers, there exist T1, T1′′, T2, T2′′ for which T1 = T1′T1′′ and T2 = T2′T2′′, in this caseT1′T2′ is always transcendental and of courseT1′ +T2′ too.
T1T2nis transcendental∀n 6=n0and T1+mT2 is transcendental∀m, in this caseT12T2n0 is transcendental, andT12T2n0 andT1+mT2are always transcen- dentals. For allT1′, T2′ prime real numbers existT1 =T′
1 2
1 andT2 =T′
1 n0
2 and
m= 1, thusT1′T2′ andT1′+T2′ are transcendentals ! II)
There are three sub-possibilities :
T1+T1′′+m(T2+T2′′)is transcendental for allm, for allT1′, T2′ transcenden- tal prime numbers, there exists T1, T1′′, T2, T2′′ for whichT1 = T1′ −T1′′ and T2 = T2′ −T2′′, in this caseT1′ +T2′ is always transcendental and of course T1′T2′ too.
T1−T1′′+m(T2−T2′′)is transcendental for allm, for allT1′, T2′ transcenden- tal prime numbers, there exists T1, T1′′, T2, T2′′ for whichT1 = T1′ +T1′′ and T2 = T2′ +T2′′, in this caseT1′ +T2′ is always transcendental and of course T1′T2′ too.
T1T2nis transcendental∀nandT1+mT2,∀m 6=m0, in this case2T1+mT2is always transcendental, for allT1′, T2′ prime real numbers existT1 = T21′ and T2 = mT2′0 andn = 1, thusT1′ +T2′ andT1′T2′ are transcendentals !
The theorem application T1′ = π and T2′ = e are transcendental prime real numbers which implies thateπ, e+π and e−π, are transcendentals for example !
πmenandπm+enare transcendental for allm, n.
Also, by the same way, we prove thatπn,em, Cnr,πnem, Cnrπn andCnrem are transcendentals for everyn, m, r!
Conclusion
Through this exposé, we have given a method to find the nature of several numbers, we have shown the nature of πnem, πnCnp, emCnp, πn, em and Cnp...
Références
[1] Alan Baker, Transcendental number theoryCambridge University Press , (1975).
